Calculating Percent Composition, (Empirical & Molecular Formulas) Unit 6 Learning Target 2 Calculating Percent Composition, (Empirical & Molecular Formulas)

What you will need (so get it out!) Periodic Table Calculator Writing Instrument Note Sheet (from last week, the one with the definitions.)

Percent Composition The percent by mass of each element in a compound. (pg. 341) Application: Percent composition= mass of element X 100 (of element) mass of compound

Let’s Practice (on back of note sheet) Start just like you would if calculating molar mass. NaCl Na 1 x 23 = 23 Cl 35 = +35 58

Next… Divide the total mass of each element by the total mass of the compound. Na 23/58 = 0.3966 Cl 35/58 = .6034

Finally… Multiply decimal by 100. Na 0.3966 x 100 39.66 % Cl 0.6034 x 100 60.34 %

One more… Try this one on your own: Mg3(PO4)2

Molar Mass Mg 3 x 24 = 72 P 2 x 31 = 62 O 8 x 16 = +128 262

Divide Mg 72/262 = 0.2749 P 62/262 = 0.2366 O 128/262 = 0.4885

Multiply by 100 Mg 0.2749 x 100 = 27.46 % P 0.2366 x100 = 23.66 % O 48.85 %

Worksheet Time Work independently or with a partner to complete the worksheet. Formulas you may need: 6) Fe3(PO4)2 7) BeN 8) KCN 9) Mn(NO3)3 10) Li3P 11) Ni2(SO4)3

Empirical Formula A formula that shows the smallest whole number ratio of the elements of a compound, and might be the same as the actual molecular formula. (pg. 344) Application: CH2O CH3OH H2O

How to solve an empirical formula problem Convert masses of elements to moles. If the given information is in percentages, assume a 100 gram sample so percents become grams. Divide moles of each element by the smallest number of moles. Round to nearest whole number unless it is close to .5 This ratio gives the subscripts for each element in the formula.

Sample Problem Methyl Acetate is a solvent commonly used in paints, inks, and adhesives. Determine the empirical formula for Methyl Acetate, which has the following chemical analysis: 48.64% Carbon, 8.16% Hydrogen, and 43.20% Oxygen.

Convert to Moles 48.64% Carbon, 8.16% Hydrogen, and 43.20% Oxygen C 48.64 % 48.64 g C x 1 mol C = 1 12 g C 4.05 mol C H 8.16 % 8.16 g H x 1 mol H = 1 1 g H 8.16 mol H O 43.20 % 43.20 g O x 1 mol O = 1 16 g O 2.70 mol O

Divide by Smallest # of Moles C 4.05 moles C = 2.70 moles 1.5 moles C 1.5 x 2 = 3 H 8.16 moles H = 3 moles H 3 x 2 = 6 O 2.70 moles O = 1 mole O 1 x 2 = 2 The number of moles of each element was multiplied by 2 to make 1.5 a whole number.

Write the formula Methyl Acetate C3H6O2

Let’s do one more! Succinic Acid is a substance produced by lichens. Chemical analysis indicates that it is composed of 40.68 g Carbon, 5.08 g Hydrogen, and 54.24 g Oxygen. Determine the empirical formula.

Convert to Moles 40.68 g Carbon, 5.08 g Hydrogen, and 54.24 g Oxygen C 40.68 g C x 1 mol C = 1 12 g C 3.39 mol C H 5.08 g 5.08 g H x 1 mol H = 1 1 g H 5.08 mol H O 54.24 g 54.24 g O x 1 mol O = 1 16 g O 3.39 mol O

Divide by the smallest # of Moles Succinic Acid= C2H3O2 C 3.39 moles C = 3.39 moles 1 moles C 1 x 2 = 2 H 5.08 moles H = 1.5 moles H 1.5 x 2 = 3 O 3.39 moles O = 1 mole O

Empirical  Molecular Formula Solved just like empirical formula with three additional steps. Find molar mass of the empirical formula. Divide molar mass of compound molar mass of empirical formula Multiply the coefficients of the empirical formula by the quotient from the previous step.

Practice Succinic Acid is a substance produced by lichens. Chemical analysis indicates that it is composed of 40.68 g Carbon, 5.08 g Hydrogen, and 54.24 g Oxygen. The molar mass of the compound is 118.1 g/mol. Determine the empirical and molecular formulas.

Sound familiar? We have already found the empirical formula. C2H3O2 Calculate the molar mass: 2(12 g/mol)+3(1 g/mol)+2(16 g/mol)= 59.0 g/mol Divide mass of molecular by mass of empirical: 118.1 g/mol = 2 59.8 g/mol 2 (C2H3O2) = C4H6O4

Check yourself 4(12 g/mol)+6(1 g/mol)+4(16 g/mol)= 118.0 g/mol

Worksheet time!